Summing combinations with repetition

Question

Given $m,n,k\in\mathbb{N}=\{1,2,...\}$, I wonder if it is possible to find a $F:\mathbb{N}^3\to \mathbb{N}$ such that

$$ \binom{m+k-1}{k}+\binom{n+k-1}{k}=\binom{F(m,n,k)+k-1}{k}. $$

EDIT: A more reasonable version of the problem (inspired by the smart observations below) is to find, for each $k$, a different function, say $F_k(m,n)$, such that $ \binom{m+k-1}{k}+\binom{n+k-1}{k}=\binom{F_k(m,n)+k-1}{k}$. For instance, for $k=1$, we have $F_1(m,n)=m+n$.

Thanks for your help!

See also https://math.stackexchange.com/a/2841171/559615.

Answer 1

Well, if such $F$ exist, then $k=1$ since for $m=n=1$ we have

$$2{k\choose k} = {F(1,1)+k-1\choose k}\implies F(1,1)=2\;\; \wedge\;\; k=1$$

(remember Pascal triangle and only possibility where 2 stands) so $$ \binom{m}{1}+\binom{n}{1}=\binom{F(m,n)}{1}, $$

So $F(m,n) = m+n$.

Answer 2

Given for ascertained that it cannot be $F(m,n)$, but the it shall be $F(m,n,k) as in your new edition of the post, then,using the Falling and Rising Factorial definition of the binomial, your identity can be rewritten as $$ \eqalign{ & \left( \matrix{ m + k - 1 \cr k \cr} \right) + \left( \matrix{ n + k - 1 \cr k \cr} \right) = \left( \matrix{ F(m,n) + k - 1 \cr k \cr} \right)\quad \Rightarrow \cr & \Rightarrow \quad {{\left( {m + k - 1} \right)^{\,\underline {\,k\,} } } \over {k!}} + {{\left( {n + k - 1} \right)^{\,\underline {\,k\,} } } \over {k!}} = {{\left( {F(m,n,k) + k - 1} \right)^{\,\underline {\,k\,} } } \over {k!}}\quad \Rightarrow \cr & \Rightarrow \quad m^{\,\overline {\,k\,} } + n^{\,\overline {\,k\,} } = F(m,n,k)^{\,\overline {\,k\,} } \cr} $$ where $$ x^{\,\overline {\,k\,} } = \left( {x + k - 1} \right)^{\,\underline {\,k\,} } = {{\Gamma (x + k)} \over {\Gamma (x)}} $$ denote respectively the Rising Factorial, the Falling Factorial and their relation with the Gamma function.

Now, if we want $F$ to be an integer, the last line looks as the analoguous for the Rising Factorial of the Fermat's Last Theorem identity, and I expect (although I am not in the position to prove it) that it might subject to the same ban.

Looking more modestly for a real $F$, then we are facing with the inversion of the Gamma function, which does not have an easy formulation.

On the other hand $z^{\,\overline {\,k\,} } $ is a polynomial in $z$ of degree $k$ $$ z^{\,\overline {\,k\,} } = z\left( {z + 1} \right) \cdots \left( {z + k - 1} \right) = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,k} \right)} {\left[ \matrix{ k \cr l \cr} \right]x^{\,l} } $$ where the square brackets indicates the Stirling Numbers of 1st kind.

The equation $$ z^{\,\overline {\,k\,} } = a $$ has always a unique non-negative solution in $z$ for non-negative values of $a$, but to find it we shall resort to numerical computation, or to the various asymptotics approximations for Gamma and Factorials.

Answer 3

The short answer is: No, such a function does not exist.

Consider the case of $k=2$. We have $\binom 42=6$. But $\binom n2\neq12$ for all $n$. This means that $F(3,3,2)$ does not exist.

It is easy to construct more similar examples based on the fact that the set of binomial coefficients $S(k)=\{\binom nk\mid n\in\Bbb{N}\}$ is not closed under addition.

May be you wanted to ask some other question?